An object's two dimensional velocity is given by #v(t) = ( sqrt(3t)-t , t^2-5t)#. What is the object's rate and direction of acceleration at #t=2 #?

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Answer 1 · 2017-04-14T09:31:29

The rate of acceleration is #=1.07ms^-2# in the direction #=248.7º#

The acceleration is the derivative of the velocity.

#a(t)=v'(t)#

#v(t) = < sqrt(3t)-t , t^2-5t> #

#a(t)= < sqrt3*1/(2sqrtt)-1,2t-5>#

When #t=2#

#a(2) = < sqrt3*1/(2sqrt2)-1,2*2-5 >#

#=< -0.39,-1 >#

The rate of acceleration is

#=||a(2)||#

#=||< -0.39,-1 >||#

#=sqrt((0.39)^2+(1)^2)#

#=1.07ms^-2#

The direction is in the 3rd quadrant

#theta=180+arctan(1/0.39)#

#=248.7º#